Question

If integer k is equal to the sum of all even multiples of 15 between 295 and 615, what is the greatest prime factor of k?

Answer 1

Answer:

The greatest prime factor of k is 11.

Step-by-step explanation:

∵  LCM( 2, 15) = 30,

Thus, the even multiple of 15 must be multiple of 30,

That is, the even multiple of 15 between 295 and 615 are

300, 330, 360,..........600

Which is an AP,

Having first term, a = 300,

Common difference, d = 30,

If n be the number terms,

Last term = $a+(n-1)d$

$=300+(n-1)30$

$=270 + 30n$

$\implies 270+30n = 600\implies 30n = 330\implies n = 11$

Hence, the sum of the all even multiple of 15 from 295 to 615,

$S_{11}=\frac{11}{2}(2(300)+(11-1)30)=\frac{11}{2}(600+300)=\frac{11}{2}(900)=11\times 450$

According to the question,

$S_{11}=k$

⇒ k = 11 × 450 = 11 × 5 × 5 × 3 × 3 × 2

Hence, the greatest prime factor of k is 11.